divide 2 x power 4 + 3 x power 3 minus 6 X + 8 by X square + 2 X + 1
Answers
I multiply this x2 by the 3x + 1 to get 3x3 + 1x2, which I put underneath:
3x^3 + 3x put down underneath the dividend
Then I change the signs, add down, and remember to carry down the "+10x – 3" from the original dividend, giving me a new bottom line of –6x2 + 10x – 3:
(3x^3 – 5x^2 + 10x – 3) – (3x^3 + 1x^2) = –6x^2 + 10x – 3
Dividing the new leading term, –6x2, by the divisor's leading term, 3x, I get –2x, so I put this on top:
(–6x^2) ÷ (3x) = –2x, which goes on top
Then I multiply –2x by 3x + 1 to get –6x2 – 2x, which I put underneath. I change signs, add down, and remember to carry down the "–3 from the dividend:
(–6x^2 + 10x –3) – (–6x^2 – 2x) = 12x – 3, which is my new last line
My new last line is "12x – 3. Dividing the new leading term of 12x by the divisor's leading term of 3x, I get +4, which I put on top. I multiply 4 by 3x + 1 to get 12x + 4. I switch signs and add down. I end up with a remainder of –7:
(12x)/(3x) = 4, 4(3x + 1) = 12x + 4, (12x – 3) – (12x + 4) = –7
This division did not come out even. What am I supposed to do with the remainder?
Think back to when you did long division with plain numbers. Sometimes there would be a remainder; for instance, if you divide 132 by 5:
132 ÷ 5: 2 × 5 = 10; put 2 on top and 10 underneath; 13 – 10 = 3; new divisor is 32; 6 × 5 = 30; put 6 on top and 30 underneath; 32 – 30 = 2; answer is 26 with remainder 2
...there is a remainder of 2. Remember how you handled that? You made a fraction, putting the remainder on top of the divisor, and wrote the answer as "twenty-six and two-fifths":
\dfrac{132}{5} = 26\,\dfrac{2}{5} = 26 + \dfrac{2}{5}
5
132
=26
5
2
=26+
5
2
Answer:
I can't understand.
so I am not able to solve.